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Sửa đề: +6x^2
x^4+4x^3+6x^2-x-10=0
=>x^4-x^3+5x^3-5x^2+11x^2-11x+10x-10=0
=>(x-1)(x^3+5x^2+11x+10)=0
=>(x-1)(x^3+2x^2+3x^2+6x+5x+10)=0
=>(x-1)(x+2)(x^2+3x+5)=0
=>x=1 hoặc x=-2
(3x-2)(4x+5)=0 (2,3x-6,9)+(0,1+2)=0
12x2+15x-8x-10=0
12x2+7x-10=0
(x-2/3)(x+5/4)=0
x=2/3 hoặc x=5/4
Vậy.........
\(x^2-2y^2-1=0\)
\(x^2-2y^2=0+1\)
\(x^2-2y^2=1\)\(\Leftrightarrow x^2=1+2y^2\)
Thấy một số chính phương khi chia cho 44 có số dư là 00 hoặc 1
- Nếu y lẻ ⇒ y2 ≡ 1(mod4)
\(\Rightarrow x^2=2y^2+1\equiv3\left(mod4\right)\) ( vô li )
Do đó y chẵn⇒ y= 2 (do y ∈ P )
Thay vào tìm được x = 3
Vậy \(\left(x,y\right)\in\left\{\left(3,2\right)\right\}\)
Bài 1:
a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)
\(\Leftrightarrow6-8x-10+2x-5=0\)
\(\Leftrightarrow-6x+11=0\)
\(\Leftrightarrow-6x=-11\)
hay \(x=\dfrac{11}{6}\)
b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)
\(\Leftrightarrow6-12x-11+3x-1=0\)
\(\Leftrightarrow-9x-6=0\)
\(\Leftrightarrow-9x=6\)
hay \(x=-\dfrac{2}{3}\)
\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\)
\(\Leftrightarrow2x-3x^2-x< 15-3x^2-6x\)
\(\Leftrightarrow x-3x^2+3x^2+6x< 15\)
\(\Leftrightarrow7x< 15\)
\(\Leftrightarrow x< \frac{15}{7}\)
\(b,\frac{1-2x}{4}-2\le\frac{1-5x}{8}+x\)
\(\Leftrightarrow\frac{1-2x}{4}-\frac{1-5x}{8}-x\le2\)
\(\Leftrightarrow\frac{2-4x}{8}-\frac{1-5x}{8}-\frac{8x}{8}\le2\)
\(\Leftrightarrow\frac{2-4x-1+5x-8x}{8}\le2\)
\(\Leftrightarrow-7x+1\le16\)
\(\Leftrightarrow-7x\le15\)
\(\Leftrightarrow x\le-\frac{15}{7}\)
\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\)
\(2x-3x^2-x< 15-3x^2-6x\)
\(2x-x+6x-3x^2+3x^2< 15\)
\(7x< 15\)
\(x< \frac{15}{7}\)
\(b,\frac{1-2x}{4}-2\le\frac{1-5x}{8}+x\)
\(2\left(1-2x\right)-16\le1-5x+8x\)
\(2-4x-16\le1+3x\)
\(-14-4x\le1+3x\)
\(-4x-3x\le1+14\)
\(-7x\le15\)
\(x\ge-\frac{15}{7}\)
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
e: =>-3x+1+6(x-1)=3(x+2)
=>-3x+1+6x-6=3x+6
=>3x-5=3x+6(vô lý)
f: =>15(2x+1)-100-2(3x+2)=8(3x-1)
=>30x+15-100-6x-4=24x-8
=>24x-89=24x-8(vô lý)